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∫(f−g) for max/min of {x|x|, x²|x|} | JEE

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Question
Let f(x)f(x) be the maximum and g(x)g(x) be the minimum of {x∣x∣, x2∣x∣}\big\{x|x|,\,x^{2}|x|\big\}. Then ∫−11[f(x)−g(x)] dx\displaystyle\int_{-1}^{1}\big[f(x)-g(x)\big]\,dx equals
A112\dfrac{1}{12}
B13\dfrac{1}{3}
C23\dfrac{2}{3}correct
D712\dfrac{7}{12}
Solution
Step 1: Compare the two functions. Since
x∣x∣−x2∣x∣≥0 ⇒ ∣x∣(x−x2)≥0,x|x|-x^{2}|x|\ge 0\ \Rightarrow\ |x|\big(x-x^{2}\big)\ge 0,
 this holds when x=0,1x=0,1 or x−x2>0⇒0<x<1x-x^{2}>0\Rightarrow 0<x<1. So x∣x∣≥x2∣x∣x|x|\ge x^{2}|x| on [0,1][0,1], and x∣x∣≤x2∣x∣x|x|\le x^{2}|x| for x<0x<0 or x>1x>1. Step 2: Read off the maximum ff and minimum gg on each region:
f(x)={x∣x∣,0≤x≤1x2∣x∣,x<0 or x>1g(x)={x2∣x∣,0≤x≤1x∣x∣,x<0 or x>1f(x)=\begin{cases}x|x|,&0\le x\le 1\\[4pt]x^{2}|x|,&x<0\ \text{or}\ x>1\end{cases}\qquad g(x)=\begin{cases}x^{2}|x|,&0\le x\le 1\\[4pt]x|x|,&x<0\ \text{or}\ x>1\end{cases}
 Hence f−g=∣x∣x∣−x2∣x∣∣f-g=\big|x|x|-x^{2}|x|\big|. On [0,1][0,1], ∣x∣=x|x|=x gives f−g=x2−x3f-g=x^{2}-x^{3}; on [−1,0][-1,0], ∣x∣=−x|x|=-x gives f−g=(x3−x2)f-g=(x^{3}-x^{2}) taken positive, i.e. x2−x3x^{2}-x^{3}. Step 3: On the whole of [−1,1][-1,1] therefore f(x)−g(x)=x2−x3f(x)-g(x)=x^{2}-x^{3}, hence
∫−11(x2−x3) dx.\int_{-1}^{1}\big(x^{2}-x^{3}\big)\,dx.
 Step 4: The term x3x^{3} is odd, so it integrates to 00; x2x^{2} is even:
∫−11(x2−x3) dx=2∫01x2 dx−0=2⋅13=23.\int_{-1}^{1}\big(x^{2}-x^{3}\big)\,dx=2\int_{0}^{1}x^{2}\,dx-0=2\cdot\dfrac13=\dfrac{2}{3}.
 Correct answer: (3)
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